x - 8 = -5Date: 06/13/2001 at 15:39:23 From: Will Sumas Subject: How to answer a question x-a=b My question is: take the problem x - 8 = -5. I know what the answer is; I can easily solve it by subtracting 5 from 8 to get 3, but our teacher wants us to show all of our work and add something to both sides. This is were I get stuck. Can you show me how to show my work? Date: 06/15/2001 at 02:07:19 From: Doctor Jeremiah Subject: Re: How to answer a question x-a=b Hi Will, Just because you know the answer doesn't mean that you can "prove" the answer is right. That's why you need to do it the way the teacher wants. If you do it that way you can "prove" that the answer must be right. The way that you deal with an equation like this is to isolate the x by itself. That means moving the -8 from the left side to the right side. But in order to modify an equation you must do the same operation to both sides of the equation. That's the trick right there. Say your equation is x + 5 = 8. In order to get the x by itself, you need to take the 5 away from the left side. The 5 is currently added to the left side, so to remove it we need to do the opposite operation. x + 5 - 5 is just x, so the opposite operation would be to subtract 5 from the left side. That will leave the x by itself. But we must do the same thing to both sides of the equation. That means subtracting 5 from BOTH sides. For the equation x + 5 = 8 this operation would look like: x + 5 = 8 x + 5 - 5 = 8 - 5 (subtract 5 from BOTH sides) x = 8 - 5 (remember that x+5-5 is just x) x = 3 Now let's try something harder like x - 8 = -5. The 8 is subtracted from the left-hand side. In order to isolate the x we must remove the -8 from the left side. x - 8 + 8 is just x, so the opposite operation would be to add 8 to the left side. But we must do it to both sides, so it would look like: x - 8 = -5 x - 8 + 8 = -5 + 8 (add 8 to BOTH sides) x = -5 + 8 (remember that x-8+8 is just x) x = 3 I hope that helps clear things up a little. If you still need help, please write back. - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/ |
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